In Eisenbud's book, he defines a minimal free resolution like in the photo. I want to ask, if $M$ is finitely generated, then would the condition $\partial(F_n)\subset mF_{n-1}$ be enough to prove that $F_i $ is finitely generated?
I tried to prove it: let $...\to F_1\to F_0\xrightarrow{\partial_0} M\to 0$ be a free resolution of $M$ s.t. $Ker(\partial_0)\subset mF_{0}$. THen $F_0/mF_0 \cong M/mM$ and hence $F_0/mF_0 = (\overline{f_1},...,\overline{f_n})$. Let $N=(f_1,...,f_n)$ and then $F_0=N+mF_0$. Here is where I'm stuck since $F_0$ is not finitely generated, thus NAK Lemma cannot be applied.
Did I miss something? Thank you for your help.
We will use the torsion functor to see the finite generation. Recall that $\operatorname{Tor}^R_i ( R/m, M)$ can be computed with any minimal free resolution $F_\bullet$ of $M$, i.e., the torsion module does not depend on which free resolution of $M$ is used. Indeed, $\operatorname{Tor}^R_i ( R/m, M)$ is the $i$th holomogy of the complex $F_\bullet \otimes_R R/m$.
The condition $\partial(F_i) \subset m F_{i-1}$ implies that $\partial \otimes R/m$ is a zero map. Hence the length of the module $\operatorname{Tor}^R_i ( R/m, M)$ is the rank of $F_i$. (Recall that if $F = \oplus_{i \in I} R$, then $F \otimes_R R/m \cong \oplus_{i \in I} R/m$ for any index set $I$. Thus, $F \otimes_R R/m$ is finitely generated iff $F$ is of finite rank.) Then the existence of a finite free resolution of $M$ proves the claim.